【LeetCode】字符串相乘

                9   9   9
          ×     6   7   8
   ----------------------
               72  72  72
           63  63  63
       54  54  54
   ----------------------
       54 117 189 135  72
   ----------------------
       54 117 189 142   2
   -----------------------
       54 117 203   2   2
   -----------------------
       54 137   3   2   2
   -----------------------
       67   7   3   2   2
   -----------------------
    6   7   7   3   2   2

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0") return "0";
        int m = num1.size();
        int n = num2.size();
        vector<int> vec;
        // i指向num1
        int start = 0;
        for(int j = n - 1; j >= 0; j--){
            // j指向num2
            int idx = start;
            for(int i = m - 1; i >= 0; i--){
                int res = (num2[j] - '0') * (num1[i] - '0');
                if(idx >= vec.size()) vec.push_back(res);
                else vec[idx] += res;                
                idx++;
            }
            start++;
        }
        string ans;
        for(int k = 0; k < vec.size() - 1; k++){
            vec[k + 1] += (vec[k] / 10);
            ans = to_string(vec[k] % 10) + ans;
        }
        ans = to_string(vec[vec.size() - 1]) + ans;
        
        return ans;
    }
};

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1 == "0" || num2 == "0") return "0";
        int m = num1.size();
        int n = num2.size();
        string ans(m + n, '0');
        for(int j = n - 1; j >= 0; j--){
            for(int i = m - 1; i >= 0; i--){
            	// 注意以下三步
            	// 个位数相乘再加上个位数，res不大于100
            	// 先计算出res，根据i、j下标，确定当前计算结果应该放置的位置
                int res = ans[i + j + 1] - '0' + (num2[j] - '0') * (num1[i] - '0');
                ans[i + j + 1] = (res % 10) + '0';
                ans[i + j] += res / 10;
            }
        }
        int k = 0;
        while(ans[k] == '0') k++;
        return ans.substr(k, m + n - k);
    }
};